1009. Product of Polynomials
1009. Product of Polynomials (25)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
This time, you are supposed to find A*B where A and B are two polynomials.
Input Specification:
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000.
Output Specification:
For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.
Sample Input2 1 2.4 0 3.2 2 2 1.5 1 0.5Sample Output
3 3 3.6 2 6.0 1 1.6
1 #include<stdio.h> 2 #include<math.h> 3 #include<stdlib.h> 4 #include<string.h> 5 6 int main() 7 { 8 double a[1010] = {}, b[1010] = {}, ans[2010] = {}; 9 int ka, kb, i, j, x, y, maxa, maxb; 10 scanf("%d", &ka); 11 scanf("%d", &x); 12 scanf("%lf", &a[x]); 13 maxa = x; 14 for(i = 1; i < ka; i++) 15 { 16 scanf("%d", &x); 17 scanf("%lf", &a[x]); 18 } 19 scanf("%d", &kb); 20 scanf("%d", &y); 21 scanf("%lf", &b[y]); 22 maxb = y; 23 for(i = 1; i < kb; i++) 24 { 25 scanf("%d", &y); 26 scanf("%lf", &b[y]); 27 } 28 int max = 0; 29 if(maxa > maxb) 30 { 31 for(i = maxa; i >= 0; i--) 32 { 33 for(j = maxb; j >= 0; j--) 34 { 35 ans[i + j] += a[i] * b[j]; 36 if((i + j) > max) 37 { 38 max = i + j; 39 } 40 } 41 } 42 } 43 else 44 { 45 for(i = maxb; i >= 0; i--) 46 { 47 for(j = maxa; j >= 0; j--) 48 { 49 ans[i + j] += b[i] * a[j]; 50 if((i + j) > max) 51 { 52 max = i + j; 53 } 54 } 55 } 56 } 57 int count = 0; 58 for(i = max; i >= 0; i--) 59 { 60 if(ans[i] != 0) 61 count++; 62 } 63 printf("%d", count); 64 for(i = max; i >= 0; i--) 65 { 66 if(ans[i] != 0) 67 { 68 printf(" %d %.1f", i, ans[i]); 69 } 70 } 71 printf("\n"); 72 return 0; 73 }